Lesson 5: Equations with Radicals

New Concept

This lesson introduces radical equations, where the variable is under a radical. You'll learn to solve for the unknown by isolating the radical and squaring both sides, and why checking for extraneous solutions is crucial.

What’s next

Get ready to apply this! You'll start with interactive examples, then move on to practice cards and challenge problems to build your skills.

Property

A radical equation is one in which the variable appears under a radical.

To Solve a Radical Equation.

1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Continue as usual to solve for the variable.

Examples

• To solve $\sqrt{y+2} = 5$, we square both sides: $(\sqrt{y+2})^2 = 5^2$, which gives $y+2 = 25$. Subtracting 2 gives the solution $y = 23$.

Property

An extraneous solution is a value that is not a solution to the original equation.

Whenever we square both sides of an equation, we must check the solutions in the original equation.

Examples

• Consider $\sqrt{x+4} = -5$. Squaring both sides gives $x+4 = 25$, so $x=21$. Checking this: $\sqrt{21+4} = \sqrt{25} = 5$, which is not $-5$. There is no solution.

Property

When squaring both sides of an equation, we must be careful to square the entire expression on either side of the equal sign. It is incorrect to square each term separately.

Examples

• To square the expression $x+5$, you calculate $(x+5)^2 = (x+5)(x+5) = x^2 + 10x + 25$.

• To solve $\sqrt{x+18} = x-2$, we square both sides. The right side becomes $(x-2)^2 = x^2-4x+4$. The equation is then $x+18 = x^2-4x+4$.

Property

To solve an equation where the variable is under a cube root, first isolate the cube root. Then, undo the cube root by cubing both sides of the equation. We do not have to check for extraneous solutions when we cube both sides of an equation.

Examples

• To solve $\sqrt[3]{y} = 4$, we cube both sides: $(\sqrt[3]{y})^3 = 4^3$, which gives the solution $y = 64$.

• Solve $2\sqrt[3]{x-1} = 6$. First, isolate the radical by dividing by 2 to get $\sqrt[3]{x-1} = 3$. Now, cube both sides: $(\sqrt[3]{x-1})^3 = 3^3$, so $x-1=27$, which gives $x=28$.

Property

To solve an equation by extraction of roots, first isolate the squared expression. Then, take the square root of both sides, remembering to include both the positive and negative roots ($\pm$). Finally, solve the resulting linear equations.

Examples

• Solve $(x+3)^2 = 49$. Take the square root of both sides: $x+3 = \pm\sqrt{49} = \pm 7$. This gives two equations: $x+3=7$ (so $x=4$) and $x+3=-7$ (so $x=-10$).

• Solve $3(y-1)^2 = 75$. First, isolate the squared part by dividing by 3: $(y-1)^2 = 25$. Take the square root: $y-1 = \pm 5$. The two solutions are $y = 1+5=6$ and $y=1-5=-4$.